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3 years ago

1. A race car moves along a circular track of radius 100m at a velocity of 25m/s. (a) What is

Physics

1 answer:

goldenfox [79]3 years ago

6 0

Given :

A race car moves along a circular track of radius 100m at a velocity of 25m/s.

To Find :

(a) What is the time taken to complete one lap of the circular track.

(b) What is the time taken for 10 laps.

Solution :

Circumference of circular track,

$C = 2\pi r\\\\C = 2\times \pi \times 100\ m\\\\C = 628.32 \ m$

a) Time taken to complete one lap is :

$t= \dfrac{628.32}{25}\ s\\\\t =25.13 \ s$

b) Time taken to complete 10 laps is :

$t_{10} = 10\times t\\\\t_{10} = 10\times 25.13\ s\\\\t_{10} = 251.3\ s$

Hence, this is the required solution.

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Explanation:

<h2>The S. I. unit of momentum is Kg. m/sec</h2>

hope it helps you 

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3 years ago

Q 24, 25, 26 i dont get them and need answers for it

Alexandra [31]

Answer:

24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)

Explanation:

24) This problem can be solved by means of the following equation.

$DU = Q-W\\$

where:

DU = internal energy difference [J]

Q = Heat transfer [J]

W = work [J]

Since there are no temperature changes the internal energy change is equal to zero

DU = 0

therefore:

$Q = W\\$

The work is equal to the heat transfered, W = 75 [J].

25) The heat transfer can be calculated by means of the following equation.

$Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]$

Q = 0.4*897*5 = 1794[J]

Work is equal to heat transfer, W = 1794[J]

26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.

$E_{p}=Q\\\\E_{p}=m*g*h$

where:

m = mass = 0.5[kg]

g = gravity = 9.81[m/s^2]

h = 1.5 [m]

$E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]$

The heat developed can be calculated by means of the following equation.

$Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]$

The number of times will be calculated as follows

n = 65/7.36

n = 8.8 (times) or 9 (times)

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3 years ago

The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi

Keith_Richards [23]

Answer:

Explanation:

a )

from lens makers formula

$\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})$

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

$\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})$

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens f = 1.8 cm

image distance v = ?

lens formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}$

$\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}$

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

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3 years ago

A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 590 m/s . Part A Part complete Wha

Kisachek [45]

Answer:

The recoil velocity is 0.354 m/s.

Explanation:

Given that,

Mass of hunter = 70 kg

Mass of bullet = 42 g = 0.042 kg

Speed of bullet = 590 m/s

We need to calculate the recoil speed of hunter

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Where, $m_{1}$ = mass of hunter

$m_{2}$ = mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

$0+0=70\times v_{1}+0.042\times590$

$v_{1}=-\dfrac{0.042\times590}{70}$

$v=-0.354\ m/s$

Hence, The recoil velocity is 0.354 m/s.

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3 years ago

For an object to be in static equilibrium, the net force must be equal to zero and and The center of gravity must be above the s

iren [92.7K]

Answer:

The normal force must equal the weight.

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3 years ago